So the first derivative w.r.t. t:
$\frac{dz}{dt} = \frac{\partial f}{\partial x} \times \frac{dx}{dt} + \frac{\partial f}{\partial f} \times \frac{dy}{dt}$
How would I find the 2nd derivative?
EDIT:
In particular there is a term in when calculating the 2nd derivative :
$$ \frac {d}{dt} ( \frac {\partial f}{\partial x}) $$
It is calculating this step that I do not understand
Thanks
Just imagine we write $\frac{\partial f(x(t),y(t))}{\partial x}=g(x(t),y(t))$ to simplify notation. You differentiate it exactly as you did before, just for a different function:
$$\frac{dg}{dt}=\frac{\partial g}{\partial x}\frac{dx}{dt}+\frac{\partial g}{\partial y}\frac{dy}{dt}$$
Now just realize that $g$ is already a derivative, so you get second derivatives: $$\frac{dg}{dt}=\frac{\partial^2 f}{\partial x^2}\frac{dx}{dt}+\frac{\partial^2 f}{\partial x\partial y}\frac{dy}{dt}$$
This step is used for parenthesised expressions between second and third line of this solution.